What is the extreme f(x)=x^4-8x^3+4 ?

The extreme f(x)=x^4-8x^3+4 is Saddle(0,4),Minimum(6,-428)

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extreme f(x)=x^4-8x^3+4

extreme points

f (x) = x^{4} - 8 x^{3} + 4

Saddle (0, 4), Minimum (6, - 428)

Solution steps

f^{'} (x) = 4 x^{3} - 24 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Decreasing

: 0 < x < 6,

Increasing

: 6 < x < \infty

Plug

x = 0

into

x^{4} - 8 x^{3} + 4 : 4

Saddle (0, 4)

Plug

x = 6

into

x^{4} - 8 x^{3} + 4 : - 428

Minimum (6, - 428)

Saddle (0, 4), Minimum (6, - 428)

y = x^{5} - x^{3}

x^{3} + 1

x 2^{- x}

f (x) = 8 x^{3} + 2 x y - 3 x^{2} + y^{2} + 1

f (x) = \frac{( x + 1 )}{x ^{2}}

What is the extreme f(x)=x^4-8x^3+4 ?
The extreme f(x)=x^4-8x^3+4 is Saddle(0,4),Minimum(6,-428)